Question:
Calculate the work done in following reaction at \(27^{\circ} \mathrm{C}\).
\[ \begin{aligned} & 4 \mathrm{SO}_{2(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 4 \mathrm{SO}_{3(\mathrm{~g})} \\ & \left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \end{aligned} \]
Calculate the work done in following reaction at \(27^{\circ} \mathrm{C}\).
\[ \begin{aligned} & 4 \mathrm{SO}_{2(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 4 \mathrm{SO}_{3(\mathrm{~g})} \\ & \left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \end{aligned} \]
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Find the change in moles of gas and use w = -delta n g R T.
Updated On: Nov 13, 2026
- 4988.4 J
- 2494.2 J
- 1247.1 J
- 3741.3 J
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
For a reaction of gases, \(w=-\Delta n_gRT\), where \(\Delta n_g\) is moles of gaseous products minus moles of gaseous reactants.
Step 2: Find change in moles
\(\Delta n_g=4-(4+2)=-2\).
Step 3: Temperature
\(T=27+273=300\text{ K}\).
Step 4: Calculation
\(w=-(-2)(8.314)(300)=+4988.4\text{ J}\).
Final Answer:
The work done is 4988.4 J. \[ \boxed{\text{(A) }\text{4988.4 J}} \]
For a reaction of gases, \(w=-\Delta n_gRT\), where \(\Delta n_g\) is moles of gaseous products minus moles of gaseous reactants.
Step 2: Find change in moles
\(\Delta n_g=4-(4+2)=-2\).
Step 3: Temperature
\(T=27+273=300\text{ K}\).
Step 4: Calculation
\(w=-(-2)(8.314)(300)=+4988.4\text{ J}\).
Final Answer:
The work done is 4988.4 J. \[ \boxed{\text{(A) }\text{4988.4 J}} \]
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